PUMaC 2014 · 数论（A 组） · 第 4 题

PUMaC 2014 — Number Theory (Division A) — Problem 4

[ 4 ] Find the sum of all positive integer x such that 3 × 2 = n − 1 for some positive integer n .

解析

[ 4 ] Find the sum of all positive integer x such that 3 × 2 = n − 1 for some positive integer n . Solution: 2 We see that n − 1 = ( n + 1)( n − 1). Clearly, n 6 = 1. Thus, we consider two cases: a b Case 1: n + 1 = 3 × 2 , n − 1 = 2 . It is clear that b ≥ a as otherwise n + 1 is at least 6 times of n − 1 which is impossible. Hence a b − a n + 1 − ( n − 1) = 2 = 2 (3 − 2 ). If a = 1, then b − a = 1 and b = 2. Hence x = a + b = 3 is a solution. When a = 0, b − a = 0 and hence x = 0. a b b a − b Case 2: n +1 = 2 , n − 1 = 3 × 2 . It is clear that a > b . Hence n +1 − ( n − 1) = 2 = 2 (2 − 3). If b = 1, then a − b = 2 and a = 3 Hence x = a + b = 4 is a solution. When b = 2, a − b is not an integer. Thus the only solutions are 3 , 4 and hence answer is 7 .